$\lim\limits_{x→\frac{\pi}{4}}\frac{4\sqrt{2}-(cosx+sinx)^5}{1-sin2x}$ is equal to :

$5\sqrt{2}$

The correct answer is Option (3) → $5\sqrt{2}$

$\lim\limits_{x→\frac{\pi}{4}}\frac{4\sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2x}=\frac{0}{0}form$

as $1-\sin 2x=\cos^2x+\sin^2x-2\cos x\sin x=(\cos x-\sin x)^2$

$≡\lim\limits_{x→\frac{\pi}{4}}\frac{4\sqrt{2}-(\cos x+\sin x)^5}{(\cos x-\sin x)^2}$

Using L'hopital's rule

$\lim\limits_{x→\frac{\pi}{4}}\frac{-5(\cos x+\sin x)^4(\cos x-\sin x)}{-2(\cos x-\sin x)(\cos x+\sin x)}$

$=\lim\limits_{x→\frac{\pi}{4}}\frac{5}{2}(\cos x+\sin x)^3=\frac{5}{2}(\sqrt{2})^3$

$=5\sqrt{2}$